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Tính nhanh: A = $\frac{2020^{3}+1}{2020^{2}-2019}$ B = $\frac{2020^{3}-1}{2020^{2}+2021}$
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Answers ( )
A=2020^3+1/2020^2-2019
A=(2020+1)(2020^2-2020+1)/2020^2-2019
A=2021.(2020^2-2019)/2020^2-2019
A=2021.
B=2020^3-1/2020^2+2021
B=(2020-1)(2020^2+2020+1)/2020^2+2021
B=2019.(2020^2+2021)/2020^2+2021
B=2019
Đáp án:
$A=2021$
$B=2019$
Giải thích các bước giải:
$A=\dfrac{2020^3+1}{2020^2-2019}$
$=\dfrac{(2020+1)(2020^2-2020+1)}{2020^2-2019}$
$=\dfrac{2021.(2020^2-2019)}{2020^2-2019}$
$=2021$
$B=\dfrac{2020^3-1}{2020^2+2021}$
$=\dfrac{(2020-1)(2020^2+2020+1}{2020^2+2021}$
$=\dfrac{2019.(2020^2+2021)}{2020^2+2021}$
$=2019$